#2081

Hard

LeetCode Problem Workspace

Sum of k-Mirror Numbers

This problem requires finding the sum of the n smallest k-mirror numbers by generating palindromes in base-k and base-10.

math enumeration

Problem Statement

A k-mirror number is a positive integer that is a palindrome in both base-10 and base-k, with no leading zeros. Given an integer k (2 <= k <= 9) and n (1 <= n <= 30), you need to return the sum of the n smallest k-mirror numbers. These numbers must satisfy the condition of being palindromes in both base-10 and base-k.

For example, when k = 2 and n = 5, the first five k-mirror numbers in base-10 are 1, 3, 5, 7, and 9. Their sum is 25. Similarly, for k = 3 and n = 7, the numbers are 1, 2, 4, 8, 121, 151, and 212, with a sum of 499.

Examples

Example 1

Input: k = 2, n = 5

Output: 25

The 5 smallest 2-mirror numbers and their representations in base-2 are listed as follows: base-10 base-2 1 1 3 11 5 101 7 111 9 1001 Their sum = 1 + 3 + 5 + 7 + 9 = 25.

Example 2

Input: k = 3, n = 7

Output: 499

The 7 smallest 3-mirror numbers are and their representations in base-3 are listed as follows: base-10 base-3 1 1 2 2 4 11 8 22 121 11111 151 12121 212 21212 Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499.

Example 3

Input: k = 7, n = 17

Output: 20379000

The 17 smallest 7-mirror numbers are: 1, 2, 3, 4, 5, 6, 8, 121, 171, 242, 292, 16561, 65656, 2137312, 4602064, 6597956, 6958596

Constraints

2 <= k <= 9
1 <= n <= 30

Solution Approach

Base-K Palindrome Generation

To efficiently find k-mirror numbers, we generate candidate numbers that are palindromes in base-10 and check if they are palindromes in base-k. By focusing on constructing numbers that are palindromes in both bases, we reduce the search space significantly.

Enumerating Palindrome Numbers

Enumerate numbers that are palindromes in base-10. For each palindrome, check if it is also a palindrome in base-k. Only numbers satisfying both conditions are valid k-mirror numbers.

Efficient Search and Summing

Once valid k-mirror numbers are identified, sum the first n numbers. As we know that n is at most 30, ensuring the algorithm generates numbers up to this size is feasible with an efficient search.

Complexity Analysis

Metric	Value
Time	O(1)
Space	O(1)

The time and space complexity of this problem is O(1), as the constraints limit the search space to a small number of k-mirror numbers (at most 30). The solution requires minimal computational resources due to the bounded nature of the input size.

What Interviewers Usually Probe

Understand the importance of generating numbers that are palindromes in both base-10 and base-k.
Recognize that brute-force checking all numbers is inefficient and a more optimized approach is needed.
Evaluate how well the candidate numbers are being generated and filtered for the k-mirror condition.

Common Pitfalls or Variants

Common pitfalls

Failing to optimize the search space and brute-forcing all potential numbers, which results in inefficiency.
Misunderstanding the dual palindrome condition for base-10 and base-k, leading to incorrect numbers being considered.
Overcomplicating the solution by not focusing on direct number generation strategies for palindromes.

Follow-up variants

Generalize the approach to handle larger values of n, beyond the limit of 30.
Consider how the algorithm might be adjusted if k were allowed to be greater than 9.
Investigate the performance of the solution for edge cases, such as when n = 1 or k = 2.

FAQ

What is a k-mirror number?

A k-mirror number is a positive integer that is a palindrome both in base-10 and base-k, with no leading zeros.

How do I check if a number is a palindrome in base-k?

To check if a number is a palindrome in base-k, convert it to base-k and verify if it reads the same forwards and backwards.

What is the time complexity of solving the 'Sum of k-Mirror Numbers' problem?

The time complexity is O(1), as the input size is small (n <= 30) and the number of valid k-mirror numbers is limited.

How does the approach for generating k-mirror numbers work?

The approach involves generating candidate palindrome numbers in base-10 and checking if they are also palindromes in base-k. Only valid k-mirror numbers are considered.

What are some common mistakes when solving this problem?

Common mistakes include inefficient brute-force methods, incorrect palindrome checks, and failure to generate valid candidate numbers efficiently.

terminal

Solution

Solution 1: Half Enumeration + Mathematics

For a k-mirror number, we can divide it into two parts: the first half and the second half. For numbers with even length, the first and second halves are exactly the same; for numbers with odd length, the first and second halves are the same, but the middle digit can be any digit.

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class Solution:
    def kMirror(self, k: int, n: int) -> int:
        def check(x: int, k: int) -> bool:
            s = []
            while x:
                s.append(x % k)
                x //= k
            return s == s[::-1]

        ans = 0
        for l in count(1):
            x = 10 ** ((l - 1) // 2)
            y = 10 ** ((l + 1) // 2)
            for i in range(x, y):
                v = i
                j = i if l % 2 == 0 else i // 10
                while j > 0:
                    v = v * 10 + j % 10
                    j //= 10
                if check(v, k):
                    ans += v
                    n -= 1
                    if n == 0:
                        return ans

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